optimization- please explain- Consider the following problem: A box with an open top is to be constructed from

Question

Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have. Let x denote the length of the side of the square being cut out. Let y denote the length of the base.

1-Write an expression for the volume

2-Use the given information to write an equation that relates the variables

3-Use part (2) to write the volume as a function of x.

4-Finish solving the problem---> V=???

Answer 1

After cutting out the 4 corner squares, each with side x, and folding up the box, the new length will be 3-2x, the new width will be 3-2x, and the new height will be x. Since Volume = length times width times height, the Volume V = (3-2x)(3-2x)(x) = 4x^3-12x^2+9x. If you know some calculus, the deritive of V with respect to x is V' = 12x^2-24x+9. Set this equal to zero and solve for x. This gives you the two possible answers x = 3/2 and x = 1/2. The first is impossible because if x = 3/2, then the length and width of the box will be zero (since 3 - 2 times 3/2 = 0). So, the answer is x = 1/2 ft = 6 inches, and plugging this value in to the equation for V gives a maximum volume of 2 cu.ft. for the box.

Answer 2

The volume of the rectangle is going to be (3 feet – 2x) for the length and (3 feet – 2x) for the width all times x (the height).

So for ALL values of x, the volume is [4x2 -12x + 9]x or 4x3 - 12x2 + 9x

To Optimize that? I yield to the one who can get the right answer.

Answer 3

The next step is to derive the function to find it maximum for 0